Help with nonlinear equation - reduction of order

**dwally89** · Jul 4th 2011, 01:34 PM

Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:
(y^3)y''=1

So far, using a substitution of p=y', I have managed to do:
(y^3)p(dp/dy)=1
p dp = 1/(y^3) dy
(p^2)/2 = -1/(2y^2) + c
p^2 = -1/(y^2) + c

Can someone please confirm if my working so far is correct or incorrect and where to go from here.

Thanks

**Also sprach Zarathustra** · Jul 4th 2011, 03:07 PM

Originally Posted by dwally89

Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:
(y^3)y''=1

So far, using a substitution of p=y', I have managed to do:
(y^3)p(dp/dy)=1
p dp = 1/(y^3) dy
(p^2)/2 = -1/(2y^2) + c
p^2 = -1/(y^2) + c

Can someone please confirm if my working so far is correct or incorrect and where to go from here.

Thanks

So far so good...

$p=\sqrt{c -\frac{1}{y^2} }$

$p=\frac{dy}{dx}$

$\frac{dy}{dx}=\sqrt{c -\frac{1}{y^2} }$

$dx=\frac{dy}{\sqrt{c-\frac{1}{y^2}}}$

Now take integral on both sides.

**dwally89** · Jul 4th 2011, 03:33 PM

Originally Posted by Also sprach Zarathustra

So far so good...

$p=\sqrt{c -\frac{1}{y^2} }$

$p=\frac{dy}{dx}$

$\frac{dy}{dx}=\sqrt{c -\frac{1}{y^2} }$

$dx=\frac{dy}{\sqrt{c-\frac{1}{y^2}}}$

Now take integral on both sides.

I managed to get to that step, but am not sure how to take the integral of the right hand side.

Thanks

**Also sprach Zarathustra** · Jul 4th 2011, 04:56 PM

Originally Posted by dwally89

I managed to get to that step, but am not sure how to take the integral of the right hand side.

Thanks

$\int\frac{dy}{\sqrt{c-\frac{1}{y^2}}}=\int \frac{y}{\sqrt{cy^2-1}}dy$

Now, substitute $u=cy^2-1$ then $du=2cy$

Try to finish...

**dwally89** · Jul 5th 2011, 01:36 AM

Originally Posted by Also sprach Zarathustra

$\int\frac{dy}{\sqrt{c-\frac{1}{y^2}}}=\int \frac{y}{\sqrt{cy^2-1}}dy$

Now, substitute $u=cy^2-1$ then $du=2cy$

Try to finish...

So:
$y=\sqrt{cx^2+\frac{1}{c}}$

which then working backwards, to check that it works gives:
$y^3=({cx^2+\frac{1}{c}})^{1.5}$
$y''=({cx^2+\frac{1}{c}})^{-1.5}$
$y^3y''=1$

Right?

**Also sprach Zarathustra** · Jul 5th 2011, 04:05 AM

Originally Posted by dwally89

So:
$y=\sqrt{cx^2+\frac{1}{c}}$

which then working backwards, to check that it works gives:
$y^3=({cx^2+\frac{1}{c}})^{1.5}$
$y''=({cx^2+\frac{1}{c}})^{-1.5}$
$y^3y''=1$

Right?

I don't think so...

Steps after taking integration on both sides:

$\frac{y\sqrt{c-\frac{1}{y^2}}}{c}+A=x+B$

Or:

$y\sqrt{c-\frac{1}{y^2}}=x+K$

Or:

$\sqrt{cy^2-1}=x+K$

Or:

$cy^2-1=(x+K)^2$

Now solve for $y$ .

**dwally89** · Jul 5th 2011, 04:19 AM

Originally Posted by Also sprach Zarathustra

I don't think so...

Steps after taking integration on both sides:

$\frac{y\sqrt{c-\frac{1}{y^2}}}{c}+A=x+B$

Or:

$y\sqrt{c-\frac{1}{y^2}}=x+K$

Or:

$\sqrt{cy^2-1}=x+K$

Or:

$cy^2-1=(x+K)^2$

Now solve for $y$ .

I have a bad habit of forgetting to add constants when integrating...

$y=\sqrt{\frac{(x+K)^2+1}{c}}$
Is that right?

**Also sprach Zarathustra** · Jul 5th 2011, 04:22 AM

Originally Posted by dwally89

$y=\sqrt{\frac{(x+K)^2+1}{c}}$
Is that right?

$y=\sqrt{\frac{(x+K)^2+1}{c}}$ or $y=-\sqrt{\frac{(x+K)^2+1}{c}}$

**dwally89** · Jul 5th 2011, 04:24 AM

Thanks

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